Euler characteristic of compact connected nontrivial Lie group is zero

Statement

Suppose $G$ is a nontrivial compact connected Lie group (?). Then, $G$ is a Space with zero Euler characteristic (?), i.e., the Euler characteristic (?) of $G$ is $0$ .

Related facts

Similar facts

Euler characteristic of odd-dimensional compact connected orientable manifold is zero

Facts used

Lefschetz fixed-point theorem
Lie group implies polyhedron: Any Lie group can be given a simplicial complex structure, and is hence a polyhedron.
Definition of Euler characteristic as the Lefschetz number of the identity map from a space to itself.
Lefschetz number is homotopy-invariant: If $f_{1}, f_{2} : X \to Y$ are homotopic maps, then the Lefschetz numbers of $f_{1}$ and $f_{2}$ are equal.

Proof

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Given: A compact connected nontrivial Lie group $G$ .

To prove: The Euler characteristic of $G$ is zero.

Proof: We denote by $e$ the identity element of $G$ .

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$G$ is a compact polyhedron, and hence a space with finitely generated homology	Fact (2)	$G$ is compact and is a Lie group		Given+Fact direct
2	Denote by $m_{e} : G \to G$ the left multiplication map by $e$ . $m_{e}$ is the identity map and the Lefschetz number of $m_{e}$ is $χ (G)$	Fact (3)	$G$ is a group, $e$ its identity element	Step (1)	[SHOW MORE] By Step (1), it makes sense to talk of the Lefschetz number of $m_{e}$ . By the definition of group and identity element, $m_{e} (x) = x$ for all $x \in G$ , so $m_{e}$ is the identity map. Fact (3) now tells us that its Lefschetz number is $χ (G)$ .
3	Let $g$ be a non-identity element of $G$ .	--	$G$ is nontrivial
4	Denote by $m_{g} : G \to G$ the left multiplication map by $g$ , a non-identity element of $g$ . In other words, $m_{g} (x) : = g x$ . Then $m_{g}$ has no fixed points in $G$	--	$G$ is a group	Step (3)	(basic group theory)
5	The Lefschetz number of $m_{g}$ is zero.	Fact (1)		Steps (1), (4)	[SHOW MORE] If the Lefschetz number were nonzero, then by Fact (1) (the Lefschetz fixed-point theorem) and Step (1) (which says that $G$ is a compact polyhedron), we see that $m_{g}$ must have a fixed point, contradicting Step (4). Hence, the Lefschetz number of $m_{g}$ must be zero.
6	$m_{e}$ and $m_{g}$ are homotopic maps		$G$ is connected and is a manifold (because it's a Lie group)		[SHOW MORE] Since $G$ is connected and a manifold, it is path-connected. Let $γ : [0, 1] \to G$ be a path with $γ (0) = e$ and $γ (1) = g$ . Consider the mapping $M : G \times [0, 1] \to G$ given by $M (x, t) = γ (t) x$ . Then $M (x, 0) = m_{e} (x)$ and $M (x, 1) = m_{g} (x)$ . Further, $M$ is continuous. Thus, $m_{e}$ and $m_{g}$ are homotopic.
7	The Lefschetz number of $m_{e}$ equals the Lefschetz number of $m_{g}$	Fact (4)	--	Step (6)	Step+Fact direct
8	The Euler characteristic of $G$ is zero (final conclusion)	--	--	Steps (2), (5), (7)	[SHOW MORE] By Step (2), the Lefschetz number of $m_{e}$ is $χ (G)$ . By Step (7), $m_{e}$ and $m_{g}$ have the same Lefschetz number. By Step (5), the Lefschetz number of $m_{g}$ is zero. Combining, we get $χ (G) = 0$ .